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Physics 18 Online
OpenStudy (anonymous):

From my previous question : At the moment the train’s speed is 45 km/h, what is the magnitude of the total accelera- tion? Answer in units of m/s2

OpenStudy (anonymous):

First convert 45 km/hr

OpenStudy (anonymous):

12.5 m/s

OpenStudy (anonymous):

heres where` the r comes from?? v^2/r

OpenStudy (anonymous):

.959

OpenStudy (anonymous):

It doesn't matter what the speed is. The earlier prompt says "slows down at a constant rate" which means the train experiences constant acceleration.

OpenStudy (anonymous):

which is?

OpenStudy (anonymous):

can you explain this further.

OpenStudy (anonymous):

You're right. I overlooked the circular part. \[a = {v^2 \over r}\]

OpenStudy (anonymous):

But that is only the centripetal component, there is still the tangential. We are asked for the total magnitude. \[a = \sqrt{\left(v^2 \over 2 \right)^2 + a_t^2}\]where \(a_t\) is the acceleration value you found in the other question.

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